Chapter 7.1, Problem 40E

To determine

To find: It needs to be determined which of the following gives the approximate velocity at $t=15$ using the Trapezoidal Rule.

Answer to Problem 40E

The approximate velocity at $t=15$ using the Trapezoidal Rule is 120 ft/sec which is option C, thus option (C) is the correct choice.

Explanation of Solution

Given information: A table is shown in the textbook where the data for the acceleration $a\left(t\right)$ of a car from 0 to 15 seconds are given and the velocity at $t=0$ is 5 ft/sec.

Formula used: The second fundamental theorem of calculus is used which says that when $F\left(x\right)$ is antiderivative of $f\left(x\right)$ then ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=F\left(b\right)-F\left(a\right)$ .

The Trapezoid rule is used which states that when $\left[a,b\right]$ is divided into n subintervals each length $\Delta x$ with endpoints at $\left\{x_0,x_1,x_2,\mathrm{...},x_n\right\}$ then one has as:

  $T_n=\frac{\Delta x}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+\mathrm{...}+2f\left(x_{n-2}\right)+2f\left(x_{n-1}\right)+f\left(x_n\right)\right]$

Calculation: As the velocity is the derivative of acceleration so one can write as ${\displaystyle \underset{0}{\overset{15}{\int }}a\left(t\right)dt}=v\left(15\right)-v\left(0\right)$ , and as velocity at $t=0$ is 0, so one has $v\left(15\right)={\displaystyle \underset{0}{\overset{15}{\int }}a\left(t\right)dt}$ .

Now further as it is known that there are 6 data points, so one has $n=6-1=5$ and width of each interval is $\Delta t=3$, thus using trapezoid rule one gets as:

  $\begin{array}{c}{T}_5=\frac{\Delta t}{2}\left[a\left(0\right)+2a\left(3\right)+2a\left(6\right)+2a\left(9\right)+2a\left(12\right)+a\left(15\right)\right]\\ =\frac{3}{2}\left[4+2\left(8\right)+2\left(6\right)+2\left(9\right)+2\left(10\right)+\left(10\right)\right]\\ =120\end{array}$

Hence the option (C) is the correct choice.